Question

Integrate the following function

$\int \frac{dx}{x( {x}^{p} + 1) }$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\displaystyle\int\rm \frac{1}{x( {x}^{p} + 1) } \: dx$

can be rewritten as

$\rm \: = \: \displaystyle\int\rm \frac{1}{x \times {x}^{p} (1 + {x}^{ - p}) } \: dx$

$\rm \: = \: \displaystyle\int\rm \frac{1}{{x}^{p + 1} \: (1 + {x}^{ - p}) } \: dx$

Now, to evaluate this integral, we use Method of Substitution.

So, Substitute

$\rm \: \: 1 + {x}^{ - p} = y$

So, on differentiating both sides w. r. t. x, we get

$\rm \: \: 0 - p{x}^{ - p - 1} = \dfrac{dy}{dx}$

$\rm \: \: - p{x}^{ - (p + 1)} = \dfrac{dy}{dx}$

$\rm\implies \:\dfrac{dx}{ {x}^{p + 1} } = - \dfrac{dy}{p}$

So, on substituting these values in above integral, we get

$\rm \: = \: - \: \dfrac{1}{p}\displaystyle\int\rm \frac{dy}{y}$

$\rm \: = \: - \: \dfrac{1}{p} \: log |y| + c$

$\rm \: = \: - \: \dfrac{1}{p} \: log |1 + {x}^{ - p} | + c$

$\rm \: = \: - \: \dfrac{1}{p} \: log \bigg|1 + \dfrac{1}{ {x}^{p} } \bigg| + c$

$\rm \: = \: - \: \dfrac{1}{p} \: log \bigg|\dfrac{ {x}^{p} + 1}{ {x}^{p} } \bigg| + c$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

ADDITIONAL INFORMATION

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

Step-by-step explanation:

Given

$\sf\int \dfrac{dx}{x( {x}^{p} + 1) }$

Solution

Refer To Attachment!!!